The potential energy of a particle of mass m is given by U(x)=E0; 0≤x≤1 0 ; x>1 λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0≤x≤1 and x>1 respectively. If the total energy of particle is 2E0,the ratio λ1λ2 will be

K.E.=2E0−E0=E0(for0≤x≤1)⇒λ1=h2mE0 K.E.=2E0(forx>1) ⇒λ2=h4mE0⇒λ1λ2=2.

The potential energy of a particle of mass m is given by U(x)=E0; 0≤x≤1 0 ; x>1 λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0≤x≤1 and x>1 respectively. If the total energy of particle is 2E0,the ratio λ1λ2 will be

K.E.=2E0−E0=E0(for0≤x≤1)⇒λ1=h2mE0 K.E.=2E0(forx>1) ⇒λ2=h4mE0⇒λ1λ2=2.

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The potential energy of a particle of mass m is given by $U (x) = \{\begin{cases} E_{0}; 0 \leq x \leq 1 \\ 0; x > 1 \end{cases}$ $λ_{1}$ and $λ_{2}$ are the de-Broglie wavelengths of the particle, when $0 \leq x \leq 1$ and $x > 1$ respectively. If the total energy of particle is $2 E_{0},$ the ratio $\frac{λ_{1}}{λ_{2}}$ will be

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$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

answer is C.

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Detailed Solution

$K . E . = 2 E_{0} - E_{0} = E_{0} (for 0 \leq x \leq 1)$
$\Rightarrow λ_{1} = \frac{h}{\sqrt{2 m E_{0}}}$
$\begin{aligned} K . E . = 2 E_{0} (for x > 1) \\ \Rightarrow λ_{2} = \frac{h}{\sqrt{4 m E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{2} . \end{aligned}$